Total electric charge on the earth,

[vorcil]

Homework Statement

The Earth can be regarded as a sphere of radius R,
the volume density of charge distributed as $$\rho = \frac{\rho_s r}{R}$$
where the density is 0 at thcce centre and rises linearly with radius until it reaches ps at the Earth's surface

i) prove that the total charge on the Earth is $$\pi \rho_s R^3$$

ii) use gauss's law to find an expression for E at the point r outside the Earth (r>R)

iii) use gauss's law to find an expression for E within the Earth's interior (r<R)

iv) write expressions for the x-, y- and z- components of E at a point outside the Earth and calculate [tex]\bf{\nabla .E }<br /> <br /> <h2>Homework Equations</h2><br /> <br /> Qenclosed = $$\int_v \rho d\tau.$$ (for a volume)<br /> <br /> $$\oint_S \bf{E} . d\bf{a} = \frac{1}{\epsilon} Qenclosed$$<br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> $$\int_v \rho \dtau$$<br /> <br /> substituting in the given rho,<br /> <br /> $$\int \rho_s \frac{r}{R} \dtau$$<br /> <br /> the outer radius r is = R so it just becomes R/R =1,<br /> <br /> $$\rho_s *1 \int \dtau$$<br /> <br /> this is where I integrate the dtau and it should give me the volume of a sphere<br /> <br /> $$\rho_s \frac{4}{3}\pi r^3$$<br /> <br /> this sort of looks like the answer I'm looking for <br /> HOWEVER I DONT KNOW HOW TO GET RID OF THE 4/3 constant!<br /> what have I done wrong?<br /> i've tried heaps of other methods, but I'm always getting some type of constant out the front[/tex]

 

[phyzguy]

The density is proportional to r, as you wrote, but you can't pull this factor of r outside the integral. You should write:
$$\int_0^R\rho(r)d\tau = \int_0^R\rho_s\frac{r}{R}4\pi r^2 dr = 4\pi\frac{\rho_s}{R}\int_0^Rr^3 dr = 4\pi \frac{\rho_s}{R} \frac{R^4}{4} = \pi\rho_sR^3$$

Does this make sense?

 

[vorcil]

The density is proportional to r, as you wrote, but you can't pull this factor of r outside the integral. You should write:
$$\int_0^R\rho(r)d\tau = \int_0^R\rho_s\frac{r}{R}4\pi r^2 dr = 4\pi\frac{\rho_s}{R}\int_0^Rr^3 dr = 4\pi \frac{\rho_s}{R} \frac{R^4}{4} = \pi\rho_sR^3$$

Does this make sense?

Yes thanks, I actually ended up doing it like this myself last night, I just didn't update it here on PF